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Natali5045456 [20]
1 year ago
8

If 8.0 g of NH4HS(s) is placed in a sealed vessel with a volume of 1.0 L and heated to 200 ∘C the reaction NH4HS(s) ⇌ NH3(g) + H

2S(g) will occur. When the system comes to equilibrium, some NH4HS(s) is still present. Which of the following changes will lead to a reduction in the amount of NH4HS(s) that is present, assuming in all cases that equilibrium is re-established following the change?
Adding more H2S(g) to the vessel
Adding more NH3(g) to the vessel
Decreasing the volume of the vessel
Adding more NH4HS(s) to the vessel
Increasing the volume of the vessel
Chemistry
1 answer:
PIT_PIT [208]1 year ago
7 0

Answer:

Increasing the volume of the vessel

Explanation:

By the Le Chatelier's principle, if a system in equilibrium suffer a variation that disturbs the equilibriu, the reaction shift in the way to minimize the pertubation and re-establish the equilibrium.

For a variation in pressure, when it increases, the reaction shift for the smallest of gas volume, and if decreases, the reaction will shift for the large gas volume. So, for the reaction given, the products have the large amount of gas, so by decreasing the pressure, more products will be formed, and the amount of NH₄HS will reduce. To decrease the pressure, we can increase the volume of the vessel: for the ideal gas equation (PV= nRT), pressure and volume are indirectly proportional.

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(1.) 163.26 g/mol, C11H17N
qwelly [4]

Answer:

75

Explanation:

4 0
2 years ago
The gas in a sealed container has an absolute pressure of 125.4 kilopascals. If the air around the container is at a pressure of
AlexFokin [52]

Answer: C. 25.6 kPa

Explanation:

The Gauge pressure is defined as the amount of pressure in a fluid that exceeds the amount of pressure in the atmosphere.

As such, the formula will be,

PG = PT – PA

Where,

PG is Gauge Pressure

PT is Absolute Pressure

PA is Atmospheric Pressure

Inputted in the formula,

PG = 125.4 - 99.8

PG = 25.6 kPa

The gauge pressure inside the container is 25.6kPa which is option C.

4 0
2 years ago
2N2H4(l) + N2O4(l) → 3N2(g) + 4H2O(g) [balanced] How many moles of N2H4 is required to produce 28.3 g of N2? Assume that all rea
JulijaS [17]

Answer: 0.67 moles of N_2H_4

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles of nitrogen}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{28.3}{28.02}=1mole

2N_2H_4(l)+N_2O_4(l)\rightarrow 3N_2(g)+4H_2O(g)

According to stoichiometry:

3 moles of N_2 is produced by 2 moles of N_2H_4

Thus 1 mole of N_2 is produced by= \frac{2}{3}\times 1=0.67moles of N_2H_4

Thus 0.67 moles of N_2H_4 are required to produce 28.3 g of N_2

6 0
2 years ago
When a 1.00 L sample of water from the surface of the Dead Sea (which is more than 400 meters below sea level and much saltier t
lisov135 [29]

Answer:

1.59mol/L

Explanation:

Data obtained from the question include:

Mass of MgCl2 = 151g

Volume of water(solvent) = 1L

Now, let us calculate the number of mole of MgCl2. This is illustrated below:

Molarity Mass of MgCl2 = 24 + (2x35.5) = 24 + 71 = 95g/mol

Mass of MgCl2 = 151g

Number of mole of MgCl2 =?

Number of mole = Mass /Molar Mass

Number of mole of MgCl2 = 151/95

Number of mole of MgCl2 = 1.59mole

Now we can calculate the molarity of MgCl2 as follow:

Mole = 1.59mole

Volume = 1L

Molarity =?

Molarity = mole /Volume

Molarity = 1.59/1

Molarity = 1.59mol/L

8 0
1 year ago
The symbol for xenon (xe) would be a part of the noble gas notation for the element
daser333 [38]

Answer:

B

Explanation:

To write the electronic configuration of an element using the preceding noble gas configuration, we simply use the noble gas in the previous period of that particular element. We do not use the noble gas of that period in which the element belongs to but the one in the preceding group.

A is wrong

This is because the noble gas in the preceding period is krypton.

In fact, the electronic configuration is [Kr] 4d105s25p3

C is wrong

The noble gas in the last period here is Radon. The electronic configuration of Radium in fact is [Rn] 7s2

D is wrong

The noble gas in the last period before the period of uranium is Radon also. In fact, the electronic configuration of the element is [Rn] 5f36d17s2

B is correct, the actual electronic configuration of the element Cesium is [Xe] 6s1

6 0
2 years ago
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